Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}2x+6y &= 4 \\ 3x+3y &= 2\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $3x = -3y+2$ Divide both sides by $3$ to isolate $x$ $x = {-y + \dfrac{2}{3}}$ Substitute this expression for $x$ in the first equation. $2({-y + \dfrac{2}{3}}) + 6y = 4$ $-2y + \dfrac{4}{3} + 6y = 4$ Simplify by combining terms, then solve for $y$ $4y + \dfrac{4}{3} = 4$ $4y = \dfrac{8}{3}$ $y = \dfrac{2}{3}$ Substitute $\dfrac{2}{3}$ for $y$ in the top equation. $2x+6( \dfrac{2}{3}) = 4$ $2x+4 = 4$ $2x = 0$ $x = 0$ The solution is $\enspace x = 0, \enspace y = \dfrac{2}{3}$.